A uniform solid cylinderal roller of mass m is being pulled on a horiz...
Given:
Mass of the cylinder, m
Force applied at the centre, F
Acceleration of the cylinder, a
The cylinder is rolling without slipping.
To find: The value of F
Analysis:
When a force is applied to the centre of a cylinder, it starts to move forward with an acceleration a. The force also creates a torque about the centre of the cylinder, which tends to rotate it.
If the cylinder is rolling without slipping, then the force of friction acts between the cylinder and the surface. This force opposes the motion of the cylinder, and its magnitude is given by:
f = ma/2
where a is the acceleration of the cylinder.
The torque due to the force F is given by:
τ = F(R/2)
where R is the radius of the cylinder.
Since the cylinder is rolling without slipping, the torque τ should be equal to the torque due to the force of friction f. Therefore,
F(R/2) = ma/2
Solving for F, we get:
F = 3ma/2R
Therefore, the value of F is 3/2 times the product of mass, acceleration, and radius of the cylinder.
Answer: The correct option is (C) 3/2 ma
A uniform solid cylinderal roller of mass m is being pulled on a horiz...
Torque is equal to product of MOI and angular accleration . rolling without slipping means alpha is equal to a/R. SOLVE THIS U GET ANSWER AS 1.5a. AS UR ANSWER
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